Middle
Secondary

Perimeter is the length of boundary around any closed figure.

The boundary of a triangle consists of its three sides. Therefore, sum of length of three sides of triangle is called its perimeter.

Units of perimeter measurement of a triangle is taken same as units of length of sides of the triangle.

The few examples of units of length of side of triangle are meter (m), centimeter (cm) etc.. So, the units of perimeter are also taken as meter (m), centimeter (cm) etc..

Perimeter of triangle \(\triangle ABC\)

In this \(\triangle ABC\), we have

length of AB = a

length of BC = b

and length of CA = c

So, perimeter of \(\triangle ABC\) is sum of length of sides AB, BC and CA

**i.e. perimeter of \(\triangle ABC\)** = \(a + b + c\)

Equilateral triangle \(\triangle ABC\) with sides length a

In equilateral triangle, where all sides are of equal length.

i.e. AB = a, BC = a and CA = a

So, **perimeter of equilateral \(\triangle ABC\)** = \(a + a + a = 3a\)

Isosceles triangle \(\triangle ABC\) with sides length a, b and a

In isosceles triangle, where any two sides are of equal length.

i.e. AB = a, BC = a and CA = b

So, **perimeter of isosceles \(\triangle ABC\)** = \(a + a + b = 2a + b\)

Example

**Find perimeter of \(\triangle ABC\) with length of its sides as given below:**

AB=2cm, BC=4cm, CA=6cm

So, perimeter of \(\triangle ABC\) = \(AB + BC + CA\)

= \(2 + 4 + 6\)

= \(12\)cm

Area is the total amount of space occupied by any closed figure.

Area is measured in square units.

The few examples of units of area of triangle are \(meter^2\) or (\(m^2\)) read as meter square, \(centimeter^2\) (\(cm^2\)) read as centimeter square etc..

Area of any triangle = \(\frac{1}{2} \times base \times height\)

Let’s understand it from following \(\triangle ABC\)

Area of triangle \(\triangle ABC\)

where, BC is base, length of BC = b

and AO is height, length of AO = h

Therefore, area of \(\triangle ABC\) = \(\frac{1}{2} \times b \times h\)

Area of equilateral triangle \(\triangle ABC\)

In above equilateral \(\triangle ABC\), where all sides of triangle are equal in length, its area is calculated as:

area of \(\triangle ABC\) = \(\frac{\sqrt 3}{4} \times a^2\)

**Let’s see how it is calculated?**

Here, in right angle \(\triangle AOC\)

\((a)^2 = (h)^2 + (\frac{a}{2})^2\) by Pythagoras theorem

\(a^2 = h^2 + \frac{a^2}{4}\)

\(a^2 – \frac{a^2}{4} = h^2\)

\(\frac{4a^2 – a^2}{4} = h^2\)

\(\frac{3a^2}{4} = h^2\)

\(\frac{\sqrt 3}{2}a = h\)

Area of \(\triangle = \frac{1}{2} \times a \times \frac{\sqrt 3}{2}a\)

Area of \(\triangle = \frac{\sqrt 3}{4}a^2\)

Area of isosceles triangle \(\triangle ABC\)

area of isosceles \(\triangle ABC = \frac{b\sqrt{4a^2-b^2}}{4}\)

**Let’s see how it is calculated?**

Here, in right angle \(\triangle AOC\)

\((a)^2 = (h)^2 + (\frac{b}{2})^2\) by Pythagoras theorem

\(a^2 = h^2 + \frac{b^2}{4}\)

\(a^2 – \frac{b^2}{4} = h^2\)

\(\frac{4a^2 – b^2}{4} = h^2\)

\(\sqrt \frac{4a^2 – b^2}{4} = h\)

\(\frac{\sqrt{4a^2-b^2}}{2} = h\)

we know, area of \(\triangle = \frac{1}{2} \times b \times h\)

\(\therefore \) area of \(\triangle ABC = \frac{1}{2} \times b \frac{\sqrt{4a^2-b^2}}{2}\)

or area of \(\triangle ABC = \frac{b\sqrt{4a^2-b^2}}{4}\)

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